題目鏈接
tag:
- Medium
question：
??Roman numerals are represented by seven different symbols: I, V, X, L, C, D and M.
| Symbol | Value |
|---|---|
| I | 1 |
| V | 5 |
| X | 10 |
| L | 50 |
| C | 100 |
| D | 500 |
| M | 1000 |
??For example, two is written as II in Roman numeral, just two one's added together. Twelve is written as, XII, which is simply X + II. The number twenty seven is written as XXVII, which is XX + V + II.
??Roman numerals are usually written largest to smallest from left to right. However, the numeral for four is not IIII. Instead, the number four is written as IV. Because the one is before the five we subtract it making four. The same principle applies to the number nine, which is written as IX. There are six instances where subtraction is used:
- I can be placed before V (5) and X (10) to make 4 and 9.
- X can be placed before L (50) and C (100) to make 40 and 90.
- C can be placed before D (500) and M (1000) to make 400 and 900.
- Given an integer, convert it to a roman numeral. Input is guaranteed to be within the range from 1 to 3999.
Example 1:
Input: 3
Output: "III"
Example 2:
Input: 4
Output: "IV"
Example 3:
Input: 9
Output: "IX"
Example 4:
Input: 58
Output: "LVIII"
Explanation: L = 50, V = 5, III = 3.
Example 5
Input: 1994
Output: "MCMXCIV"
Explanation: M = 1000, CM = 900, XC = 90 and IV = 4.
之前那篇文章寫的是羅馬數(shù)字轉(zhuǎn)化成整數(shù) Roman to Integer萌腿， 這次變成了整數(shù)轉(zhuǎn)化成羅馬數(shù)字，基本算法還是一樣。由于題目中限定了輸入數(shù)字的范圍(1 - 3999), 使得題目變得簡單了不少。
例如整數(shù) 1437 的羅馬數(shù)字為 MCDXXXVII斩例， 我們不難發(fā)現(xiàn)惶我，千位，百位等曼，十位和個位上的數(shù)分別用羅馬數(shù)字表示了裂允。 1000 - M, 400 - CD, 30 - XXX, 7 - VII损离。所以我們要做的就是用取商法分別提取各個位上的數(shù)字，然后分別表示出來：
100 - C
200 - CC
300 - CCC
400 - CD
500 - D
600 - DC
700 - DCC
800 - DCCC
900 - CM
解法一：
我們可以分為四類绝编，100到300一類草冈，400一類，500到800一類瓮增，900最后一類怎棱。每一位上的情況都是類似的，代碼如下：
class Solution {
public:
string intToRoman(int num) {
string res = "";
vector<char> roman{'M', 'D', 'C', 'L', 'X', 'V', 'I'};
vector<int> value{1000, 500, 100, 50, 10, 5, 1};
for (int n = 0; n < 7; n += 2) {
int x = num / value[n];
if (x < 4) {
for (int i = 1; i <= x; ++i) res += roman[n];
} else if (x == 4) {
res = res + roman[n] + roman[n - 1];
} else if (x > 4 && x < 9) {
res += roman[n - 1];
for (int i = 6; i <= x; ++i) res += roman[n];
} else if (x == 9) {
res = res + roman[n] + roman[n - 2];
}
num %= value[n];
}
return res;
}
};
解法二：
本題由于限制了輸入數(shù)字范圍這一特殊性绷跑，故而還有一種利用貪婪算法的解法拳恋，建立一個數(shù)表，每次通過查表找出當(dāng)前最大的數(shù)砸捏，減去再繼續(xù)查表谬运。參見代碼如下：
class Solution {
public:
string intToRoman(int num) {
string res = "";
vector<int> val{1000, 900, 500, 400, 100, 90, 50, 40, 10, 9, 5, 4, 1};
vector<string> str{"M", "CM", "D", "CD", "C", "XC", "L", "XL", "X", "IX", "V", "IV", "I"};
for (int i = 0; i < val.size(); ++i) {
while (num >= val[i]) {
num -= val[i];
res += str[i];
}
}
return res;
}
};
解法三：
下面這種方法個人感覺屬于比較投機(jī)取巧的方法，把所有的情況都列了出來垦藏，然后直接按位查表梆暖，O(1)的時間復(fù)雜度啊，參見代碼如下：
class Solution {
public:
string intToRoman(int num) {
string res = "";
vector<string> v1{"", "M", "MM", "MMM"};
vector<string> v2{"", "C", "CC", "CCC", "CD", "D", "DC", "DCC", "DCCC", "CM"};
vector<string> v3{"", "X", "XX", "XXX", "XL", "L", "LX", "LXX", "LXXX", "XC"};
vector<string> v4{"", "I", "II", "III", "IV", "V", "VI", "VII", "VIII", "IX"};
return v1[num / 1000] + v2[(num % 1000) / 100] + v3[(num % 100) / 10] + v4[num % 10];
}
};
